You need to specify in more detail the output of the LED.   What is the far-field radiation pattern - is it lensed, or does it act as a Lambertian source?  How large is the emitting area - is your surface sufficiently far from the LED to avoid near field effects (more likely to be a concern with a lensed LED or extended source).

For a sufficiently small emitter, a useful parameter is the radiant intensity - the power emitted per solid angle.

If we assume an un-lensed Lambertian emitter, radiant intensity is related to radiant flux by:  Ie = Φe / π

Solid angle subtended by surface area A, distance r from source:  Ω = A/r2

Radiant flux intercepted by surface A is:   ΦS = Ie Ω = Φe A / (π r2)

Irradiance at surface A:   Ee = ΦS / A = Φe / (π r2) = 1.4 / (72π) W cm-2 = 9.1 mW cm-2.

Radiant exposure after time t :    He = Ee t = 32.7 J cm-2 after 1 hour.

https://en.wikipedia.org/wiki/Radiant_intensity

Dear Alan Robinson

Many thanks for all the detailed information you have described and apologies for late reply, I did not visit Researchgate for some time.

1. Assuming the area of led output to be around 63 cm2 at distance 7cm from source. How can we include that information to come up with overall mj/cm2 value?  Please find attached, the radiation pattern of the led.

2. I have tried to add numbers, from a research article ,into the equations you have mentioned but results did not match with the article readings. Please find calculation below:

Radiant flux=5.2mW,     Distance between source and sample= 20mm

Duration of exposure= 150s

My calculation result was 60 mJ cm-2 whereas in the article their result was 5 mJ cm-2.

Once again thank you very much for your help. It means a lot to me.

Article Kinetics of inactivation and photoreactivation of Escherichi...

The calculation I supplied was for a small Lambertian source.  The dimensions of your lamp are comparable with the sample distance, and it is partially collimated, so not Lambertian.

What is the shape of your lamp?  If it is circular, the radius is approximately 4.5 cm, so the angle subtended by the edge of the lamp at a sample in the centre of the beam will be at 33 degrees.  According to your polar diagram, less than 20% of the peak radiant intensity is emitted at this angle.  If the lamp has a rectangular exit face, an even smaller fraction will be received from the corners.  We also need to take account of the inverse square law fall-off with distance, and a cosine inclination factor.

In broad terms, your lamp concentrates its light into a cone with half-angle 30 degrees, so the radiant intensity on-axis is potentially 4x higher than from a Lambertian source.  However, at 7 cm from the lamp, this improvement applies only to light emitted near the centre of the exit face of the lamp, and you cannot expect a 4x enhancement at your sample.

Is the radiant intensity and polar diagram uniform across the exit face of the lamp?  Is it constructed from a single relatively small emitter with a large lens or mirror to partially collimate the beam?  In this case, where is the beam focussed, and is the radiosity (W m-2) uniform across the exit face?  Alternatively does it contain multiple smaller emitters, each with its own lens?  If this case, are they accurately parallel, or are they oriented to optimise angular coverage or spatial uniformity at some specific distance from the lamp?

Regarding the paper by Zhou et. al, they also have an extended collimated source, so the simple calculations in my first response do not apply.

Their total flux is 5.2 mW and the intensity 0.035 mW cm-2 close to the lamp. This is consistent with uniform irradiance over a circular area of 149 cm2 .  13.7 cm diameter (or non-uniform centre-weighted irradiance over a larger area) seems reasonable if it is designed to accommodate Petri dishes with 9 cm internal diameter.

If you know how the radiance of your source varies across the surface of the LED, and how the polar emission varies, you can calculate the contribution from each part of the lamp - or from each indvidual LED if it contains multiple lensed emitters.

How accurately do you need to calculate the exposure?  Would it be simpler to measure the irradiance at the sample with a calibrated detector?  Alternatively, measure the relative intensity over an extended surface, and normalise the intensity at the sample using the known radiant flux from the LED.  This could be done with an inexpensive silicon photodiode.

As a crude estimate, if the effective area illuminated by the LED increases from 63 cm2 at the exit face to 100 cm2 at 7 cm separation, the irradiance will decrease from 22 mW cm-2 to approximately 14 mW cm-2 at 7 cm.

https://en.wikipedia.org/wiki/Luminance

https://en.wikipedia.org/wiki/Radiance

Dear Alan Robinson

Thank you very much for taking the time in writing a detailed reply. Apologies for late reply from my side.

As per your previous answer, I looked through some literature to figure out the correct method for calculation. I found some material, for time being the results look accurate. Hope remains the same in future. 

Once again thank you very much for you support, it has helped me a lot.

I need to radiate 1.5mj/sq cm to a surface of 144 sq inch @ 929 sq cm. What would be the wattage of the UV bulb required? Thanks.

Soh Che Kin said "I need to radiate 1.5mj/sq cm to a surface of 144 sq inch @ 929 sq cm. What would be the wattage of the UV bulb required?"

It depends. Not enough information.

Have you read any of the answers from 2017, or the Wikipedia articles linked in them?

"Wattage" is an ambiguous term. Are you referring to radiant flux? "Wattage" is often used instead to mean the electrical power consumption of a lamp.

• What kind of bulb do you intend to use? Is it a fluorescent tube which radiates in all directions, a light emitting diode (LED) with a Lambertian radiation pattern, or a lensed LED with a more directional output?
• Does the source consist of a single small emitting area, an extended uniformly emitting surface, or an array of individual emitters?
• What is the shape of the surface to be irradiated?
• How far from the surface is the lamp located?
• Is 1.5 mJ cm-2 an average over the surface, or a lower limit? Are there upper and lower limits on the variation of irradiance across the surface?

Depending on the source and its size, radiant intensity may be be a more useful metric than total radiant flux emitted - or more likely one of these together with the angular distribution of radiant intensity. If you require the electrical power consumption, this can vary enormously with the type of source you select.

Thanks Alan. Here are the specification that i got from the manufacturer today:

1. Bulb is Excimer filtered 222nm 2W

2. Bulb radiate 61uW/cm2 at 10cm distance.

3. Surface is is 929 cm2, demand a minimum 1.5mJ/cm2 average all over the target surface for 25 minutes.

4. Surface type is HEPA14 filter.

Will this choice of 2W bulb and setup siffice. Thanks.

Soh Che Kin - we still don't have full details, but I can make some assumptions for a reasonable estimate.

Assume:

• The emitting region of the bulb is small compared with the dimensions of the target.
• The bulb emits uniformly in all directions

Irradiance at distance r0 = 10 cm from source E0 = 61 µW cm-2

Radiant flux from bulb = 4 π r02 E0 = 0.077 W

Radiant exposure 10 cm from source after 25 minutes (time t = 25 x 60 =1500 s):

H = E t = 92 mJ cm-2

Minimum exposure required is 1.5 mJ cm-2, or 1 µW cm-2 for 25 minutes.

This is more than you need, but most of the surface is more than 10 cm from the source. Two factors reduce the exposure: an inverse square dependence on distance from the source, and a cosine inclination factor if the local angle of incidence is less than 90°.

What is the size and shape of the filter? Assume a 30.5 cm x 30.5 cm square.

If the bulb is height h above the centre of the square, the half-diagonal distance from centre to a corner of the square is d = 30.5 /√2 cm = 21.6 cm

Distance from source to corner of square, r = √(h2 + d2)

Irradiance at corner of square E = E0 cos θ r02 / r2 = E0 r02 h / (h2 + d2)3/2

At 10 cm height, the irradiance in the corners is 4.5 µW cm-2, which is sufficient, but much lower than directly under the bulb in the centre of the square. Increasing the height to 14 cm halves the irradiance in the centre, but increases the corner irradiance to 5 µW cm-2. In this regime the cosine inclination factor changes more rapidly than the inverse square dependence.

For a more uniform exposure over the surface of the filter, you can increase the height of the bulb above the surface to 73 cm.

Alternatively, from 40 cm height, the irradiance is 2.6 µW cm-2 in the corners, 3.8 µW cm-2 in the centre, and you can deliver the specified 1.5 mJ cm-2 in a little less than 10 minutes.

Dear Alan, all your assumptions are spot-on. icing on the cake was your radiance vs height chart, this helps a lot. Execution of my project will be based on your workings.

Appreciate it much and very generous of you to share.

God bless.

Cheers!